Question: If the following infinite geometric series converges, find its sum. $\sum\limits_{k=0}^{{\infty}}{{{1\left(-\dfrac13\right)^{k}}}} = $ Choose 1 answer: Choose 1 answer: (Choice A) A $0.75$ (Choice B) B $ 1.5 $ (Choice C) C $ 3 $ (Choice D) D $ 3.5$ (Choice E) E The series does not converge.
Answer: Overview First we need to figure out whether the geometric series sums to a specific value (converges) or sums to positive or negative infinity (diverges). Then, if the series converges, we'll compute it using the formula $S = \dfrac{a_1}{1-r}$ where $a_1$ is the first term and $r$ is the number we multiply by to get from one term to the next (i.e., the common ratio between the terms). [Where does this formula come from?] Step 1: Find $r$ In general, an infinite geometric series is $ar^0 + ar^1 + ar^2 +...$ where $r$ is the number we multiply by to go from one term to the next. See if you can find $r$ yourself (If you're stuck, click "Explain" below the problem): What is $r$ in the series in the question? For reference, the series is $\sum\limits_{k=0}^{{\infty}}{{{1\left(-\dfrac13\right)^{k}}}}$, which expands to $1 -\dfrac13+ \dfrac19 -...$ Choose 1 answer: Choose 1 answer: (Choice A) A $r=-9$ (Choice B) B $r=-1$ (Choice C) C $r=-\dfrac13$ Check Nice! Now that we know $r$, we should be able to say whether or not the series converges. Step 2: Determine if the series converges There are two basic rules for infinite geometric series: The series converges if $|r| < 1$. The series diverges if $|r| \ge 1$. [How do we know these rules are true?] Try it yourself: Does the series in the question converge? For reference, the series is $\sum\limits_{k=0}^{{\infty}}{{{1\left(-\dfrac13\right)^{k}}}}$, which expands to $1 -\dfrac13+ \dfrac19 -...$ Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No Check Step 3: Apply the formula Now that we've found $r$, we can compute the infinite geometric series $S = 1 -\dfrac13 + \dfrac19 -...$ using the formula $S = \dfrac{a_1}{1-r}$ where $a_1$ (the first term) is ${1}$ and $r$ is ${-\dfrac13}$ (see previous hint). Substitute and simplify: $\begin{aligned} S &= \dfrac{a_1}{1-r} \\\\\\\\ &= \dfrac{{1}}{1-\left({-\dfrac13}\right)} \\\\\\\\ &=0.75 \end{aligned}$ The answer The infinite geometric series converges to $ 0.75 $.